is it needed the integration by parts?integral of the  function e^x*cos^2x use integration by parts,or what else?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Yes, it uses the method!

Int f(x)dx = Int e^x*cos^2x dx

We'll substitute the square of cosine of function, by the formula:

(cos x)^2 = (1 + cos 2x)/2

We'll re-write the integral:

Int e^x*cos^2x dx = Int e^x*(1 + cos 2x)dx/2

We'll apply the additive property of indefinite integrals:

Int e^x*(1 + cos 2x)dx/2 = Int e^xdx/2 + Int (e^x*cos 2x)dx/2

We'll note Int e^xdx/2 = I1

Int (e^x*cos 2x)dx/2 = I2

I1 = e^x/2 + C (1)

We'll solve I2 by parts:

u = cos 2x => du = -2 sin 2x

dv = e^xdx => v = e^x

I2 = u*v - Int vdu

I2 = e^x*cos 2x + 2Int e^x*sin 2x dx

We'll solve 2Int e^x*sin 2x dx by parts:

u = sin 2x => du = 2 cos 2x

dv = e^xdx => v = e^x

2Int e^x*sin 2x = 2e^x*sin 2x - 4 Int e^x*cos 2xdx

But Int e^x*cos 2xdx = I2

I2 = e^x*cos 2x + 2e^x*sin 2x - 4 I2

We'll add 4I2 both sides:

5I2 = e^x*cos 2x + 2e^x*sin 2x

I2 = (e^x*cos 2x + 2e^x*sin 2x)/5 + C (2)

Int e^x*(1 + cos 2x)dx/2 = (1) + (2)

Int e^x*(1 + cos 2x)dx/2 = e^x/2 + (e^x*cos 2x + 2e^x*sin 2x)/10 + C

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